Second-order system simulation in Simulink - Step-by-Step Execution

Concept Flow - Second-order system simulation

Define system parameters: m, c, k

↓

Formulate differential equation

↓

Convert to state-space form

↓

Build Simulink model blocks

↓

Set simulation time and input

↓

Run simulation

↓

Observe output: displacement, velocity

This flow shows how to simulate a second-order system by defining parameters, modeling equations, building the Simulink model, running the simulation, and observing results.

Execution Sample

Simulink

m = 1; c = 0.5; k = 2;
A = [0 1; -k/m -c/m];
B = [0; 1/m];
C = [1 0];
D = 0;
sys = ss(A,B,C,D);
step(sys);

This code defines a second-order system's state-space model and simulates its step response.

Execution Table

Step	Action	Variable/Block	Value/Result	Notes
1	Define mass (m)	m	1	Mass set to 1 kg
2	Define damping coefficient (c)	c	0.5	Damping set to 0.5 Ns/m
3	Define spring constant (k)	k	2	Spring stiffness set to 2 N/m
4	Form A matrix	A	[[0,1],[-2,-0.5]]	State matrix for system dynamics
5	Form B matrix	B	[[0],[1]]	Input matrix for force input
6	Form C matrix	C	[1,0]	Output matrix to observe displacement
7	Form D matrix	D	0	Direct transmission matrix
8	Create state-space system	sys	ss object	System ready for simulation
9	Run step response simulation	step(sys)	Plot of displacement vs time	Shows system response to step input
10	Simulation ends	-	-	Step response completed

💡 Simulation ends after step response is fully computed and plotted

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	Final
m	undefined	1	1	1	1	1	1	1	1	1
c	undefined	undefined	0.5	0.5	0.5	0.5	0.5	0.5	0.5	0.5
k	undefined	undefined	undefined	2	2	2	2	2	2	2
A	undefined	undefined	undefined	undefined	[[0,1],[-2,-0.5]]	[[0,1],[-2,-0.5]]	[[0,1],[-2,-0.5]]	[[0,1],[-2,-0.5]]	[[0,1],[-2,-0.5]]	[[0,1],[-2,-0.5]]
B	undefined	undefined	undefined	undefined	undefined	[[0],[1]]	[[0],[1]]	[[0],[1]]	[[0],[1]]	[[0],[1]]
C	undefined	undefined	undefined	undefined	undefined	undefined	[1,0]	[1,0]	[1,0]	[1,0]
D	undefined	undefined	undefined	undefined	undefined	undefined	undefined	0	0	0
sys	undefined	undefined	undefined	undefined	undefined	undefined	undefined	undefined	ss object	ss object

Key Moments - 2 Insights

Why do we convert the second-order differential equation into state-space form?

Because Simulink works well with state-space models, which represent the system as first-order equations. This is shown in execution_table steps 4 to 8 where matrices A, B, C, D are formed and combined into the state-space system.

What does the step response simulation show?

It shows how the system's output (displacement) changes over time when a sudden force (step input) is applied. This is the output in execution_table step 9.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of matrix A after step 4?

A [[0,0],[0,0]]

B [[1,0],[0,1]]

C [[0,1],[-2,-0.5]]

D [[2,0],[0,0.5]]

💡 Hint

Refer to execution_table row 4 under 'Value/Result' column.

At which step is the state-space system object 'sys' created?

A Step 6

B Step 8

C Step 5

D Step 9

💡 Hint

Check execution_table rows 7 to 9 for when 'sys' variable is assigned.

If the damping coefficient c was increased, which matrix would change in the execution table?

A Matrix A

B Matrix C

C Matrix B

D Matrix D

💡 Hint

Look at how matrix A is formed in step 4; it depends on c.

Concept Snapshot

Second-order system simulation steps:
1. Define parameters m, c, k
2. Convert to state-space matrices A, B, C, D
3. Build system with ss(A,B,C,D)
4. Run step response with step(sys)
5. Observe output displacement over time
Use state-space for easy simulation in Simulink.

Full Transcript

To simulate a second-order system, first define the physical parameters: mass (m), damping (c), and spring constant (k). Then, write the system as a state-space model using matrices A, B, C, and D. In Simulink, build the model using these matrices. Run the simulation with a step input to see how the system responds over time. The output typically shows displacement or velocity. This process helps understand system dynamics visually and numerically.

Practice

1. What does the damping ratio (ζ) affect in a second-order system simulation in Simulink?

easy

A. The color of the simulation plot

B. The speed and overshoot of the system's response

C. The size of the Simulink blocks

D. The input signal amplitude

Solution

1. Step 1: Understand damping ratio role

   The damping ratio controls how oscillations decay in the system response.

2. Step 2: Connect damping ratio to response behavior

   A higher damping ratio reduces overshoot and speeds settling time, affecting speed and overshoot.

3. Final Answer:

   The speed and overshoot of the system's response -> Option B

4. Quick Check:

   Damping ratio = speed and overshoot control [OK]

Hint: Damping ratio controls oscillation and overshoot [OK]

Common Mistakes:

• Confusing damping ratio with input amplitude
• Thinking damping ratio changes block size
• Assuming damping ratio affects plot color

2. Which of the following is the correct way to represent a second-order transfer function in Simulink's Transfer Function block?

easy

A. Numerator: [1 0 0], Denominator: [1 2*zeta*wn wn]

B. Numerator: [1 2*zeta*wn wn^2], Denominator: [1]

C. Numerator: [1], Denominator: [1 2*zeta*wn wn^2]

D. Numerator: [wn^2], Denominator: [1 2*zeta*wn 1]

Solution

1. Step 1: Recall second-order transfer function form

   The standard form is numerator = [1], denominator = [1 2*zeta*wn wn^2].

2. Step 2: Match options to standard form

   Numerator: [1], Denominator: [1 2*zeta*wn wn^2] matches the correct numerator and denominator coefficients for a second-order system.

3. Final Answer:

   Numerator: [1], Denominator: [1 2*zeta*wn wn^2] -> Option C

4. Quick Check:

   Standard form = Numerator: [1], Denominator: [1 2*zeta*wn wn^2] [OK]

Hint: Denominator has 1, 2*zeta*wn, wn squared [OK]

Common Mistakes:

• Swapping numerator and denominator
• Using wrong powers for wn
• Missing damping term 2*zeta*wn

3. Given the Simulink transfer function block with numerator [1] and denominator [1 4 4], what is the natural frequency (wn) and damping ratio (ζ) of the system?

medium

A. wn=2, ζ=1

B. wn=4, ζ=2

C. wn=1, ζ=4

D. wn=2, ζ=0.5

Solution

1. Step 1: Identify denominator coefficients

   Denominator is [1 4 4], which matches [1 2*zeta*wn wn^2].

2. Step 2: Calculate wn and ζ from coefficients

   wn^2 = 4 -> wn = 2; 2*zeta*wn = 4 -> ζ = 4/(2*2) = 1.

3. Final Answer:

   wn=2, ζ=1 -> Option A

4. Quick Check:

   wn=2, ζ=1 from denominator [OK]

Hint: wn is sqrt of last denominator term [OK]

Common Mistakes:

• Confusing wn^2 with wn
• Dividing damping term incorrectly
• Ignoring coefficient order

4. You created a second-order system in Simulink with numerator [1] and denominator [1 0 4], but the simulation shows no oscillations. What is the likely error?

medium

A. The damping ratio is zero, causing no oscillations

B. The input signal is missing

C. The numerator should be [1 0 4]

D. The denominator coefficients are incorrect for oscillations

Solution

1. Step 1: Analyze denominator coefficients

   Denominator [1 0 4] means 2*zeta*wn = 0, so damping ratio ζ = 0.

2. Step 2: Understand effect on oscillations

   Zero damping ratio means undamped oscillations, but the system denominator form is incorrect for natural frequency 2 (since wn^2=4).

3. Step 3: Identify correct denominator form

   Correct denominator should be [1 0 4] only if system is undamped, but zero middle term can cause simulation issues.

4. Final Answer:

   The denominator coefficients are incorrect for oscillations -> Option D

5. Quick Check:

   Incorrect denominator = no oscillations [OK]

Hint: Check denominator matches 1, 2*zeta*wn, wn^2 [OK]

Common Mistakes:

• Assuming zero damping means no oscillations
• Swapping numerator and denominator
• Ignoring input signal presence

5. You want to simulate a second-order system with a natural frequency of 5 rad/s and a damping ratio of 0.3 in Simulink. What should be the denominator coefficients for the Transfer Function block?

hard

A. [1 3 25]

B. [1 3 5]

C. [1 3 15]

D. [1 3 30]

Solution

1. Step 1: Calculate denominator coefficients

   Given wn=5, ζ=0.3, calculate 2*ζ*wn = 2*0.3*5 = 3 and wn^2 = 25.

2. Step 2: Form denominator array

   Denominator coefficients are [1, 3, 25].

3. Final Answer:

   [1 3 25] -> Option A

4. Quick Check:

   2*zeta*wn=3, wn^2=25 [OK]

Hint: Multiply 2*damping*wn and square wn for denominator [OK]

Common Mistakes:

• Using wn instead of wn squared
• Mixing damping ratio with wn
• Incorrect multiplication order